1608. Special Array With X Elements Greater Than or Equal X LeetCode Solution
In this guide, you will get 1608. Special Array With X Elements Greater Than or Equal X LeetCode Solution with the best time and space complexity. The solution to Special Array With X Elements Greater Than or Equal X problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Special Array With X Elements Greater Than or Equal X
You are given an array nums of non-negative integers. nums is considered special if there exists a number x such that there are exactly x numbers in nums that are greater than or equal to x.
Notice that x does not have to be an element in nums.
Return x if the array is special, otherwise, return -1. It can be proven that if nums is special, the value for x is unique.
Example 1:
Input: nums = [3,5]
Output: 2
Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.
Example 2:
Input: nums = [0,0]
Output: -1
Explanation: No numbers fit the criteria for x.
If x = 0, there should be 0 numbers >= x, but there are 2.
If x = 1, there should be 1 number >= x, but there are 0.
If x = 2, there should be 2 numbers >= x, but there are 0.
x cannot be greater since there are only 2 numbers in nums.
Input: nums = [0,4,3,0,4]
Output: 3
Explanation: There are 3 values that are greater than or equal to 3.
Constraints:
1 <= nums.length <= 100
0 <= nums[i] <= 1000
Complexity Analysis
Time Complexity: O(\texttt{sort})
Space Complexity: O(\texttt{sort})
1608. Special Array With X Elements Greater Than or Equal X LeetCode Solution in C++
class Solution {
public:
int specialArray(vector<int>& nums) {
ranges::sort(nums);
if (nums[0] >= nums.size())
return nums.size();
for (int i = 1; i < nums.size(); ++i) {
const int count = nums.size() - i;
if (nums[i - 1] < count && nums[i] >= count)
return count;
}
return -1;
}
}; /* code provided by PROGIEZ */
1608. Special Array With X Elements Greater Than or Equal X LeetCode Solution in Java
class Solution {
public int specialArray(int[] nums) {
Arrays.sort(nums);
if (nums[0] >= nums.length)
return nums.length;
for (int i = 1; i < nums.length; ++i) {
final int count = nums.length - i;
if (nums[i - 1] < count && nums[i] >= count)
return count;
}
return -1;
}
} // code provided by PROGIEZ
1608. Special Array With X Elements Greater Than or Equal X LeetCode Solution in Python
class Solution:
def specialArray(self, nums: list[int]) -> int:
nums.sort()
if nums[0] >= len(nums):
return len(nums)
for i, (a, b) in enumerate(itertools.pairwise(nums)):
count = len(nums) - i - 1
if a < count and b >= count:
return count
return -1 # code by PROGIEZ
